Equation of tangent drawn to the parabola will be in the form :
y = mx + (a/m)
Equation of tangent drawn to the ellipse will be in the form :
y = mx ± √a^{2} m^{2} + b^{2}
Equation of tangent drawn to the hyperbola will be in the form
y = mx ± √a^{2} m^{2} + b^{2}
Point of contact of parabola :
Point of contact of Ellipse :
Point of contact of Hyperbola :
Problem 1 :
Find the equations of the tangents to the parabola y^{2} = 5x from the point (5, 13). Also find the points of contact.
Solution :
The equation of the parabola is y^{2} = 5x.
y^{2} = 4ax
5x = 4ax
a = 5/4
Equation of the tangent y = mx + a/m
y = mx + 5/4m --- (1)
Passes through the points (5, 13).
13 = 5m + 5/4m
13 = (20m^{2} + 5)/4m
52m = 20m^{2} + 5
20m^{2} - 52m + 5 = 0
20m^{2} - 2m - 50m + 5 = 0
2m(10m - 1) - 5(10m - 1) = 0
(2m - 5) (10m - 1) = 0
2m - 5 = 0 and 10m - 1 = 0
2m = 5 and 10m = 1
m = 5/2 and m = 1/10
m = 5/2 substitute the equation (1).
m = 1/10 substitute the equation (1).
So, the points of contact are (1/5, 1), (125, 25).
Problem 2 :
Find the equation of the two tangents that can be drawn
From the point (2, -3) to the parabola y^{2} = 4x.
Solution :
(i) Given, y^{2} = 4x
y^{2} = 4ax
4x = 4ax
a = 1
Passes through the points (2, -3).
Equation of the tangent to the parabola will be of the form y = mx + 1/m.
-3 = 2m + 1/m
-3 = (2m^{2} + 1)/m
-3m = 2m^{2} + 1
2m^{2} - 3m + 1 = 0
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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